Solve (-3-4i)(2+4i)

$$( - 3 - 4 i ) ( 2 + 4 i )$$

Evaluate

$$10-20i$$

Solution Steps

Multiply complex numbers -3-4i and 2+4i like you multiply binomials.

$$-3\times 2-3\times \left(4i\right)-4i\times 2-4\times 4i^{2}$$

By definition, i^{2} is -1.

$$-3\times 2-3\times \left(4i\right)-4i\times 2-4\times 4\left(-1\right)$$

Do the multiplications.

$$-6-12i-8i+16$$

Combine the real and imaginary parts.

$$-6+16+\left(-12-8\right)i$$

Do the additions.

$$10-20i$$

Real Part

$$10$$

Solution Steps

Multiply complex numbers -3-4i and 2+4i like you multiply binomials.

$$Re(-3\times 2-3\times \left(4i\right)-4i\times 2-4\times 4i^{2})$$

By definition, i^{2} is -1.

$$Re(-3\times 2-3\times \left(4i\right)-4i\times 2-4\times 4\left(-1\right))$$

Do the multiplications in -3\times 2-3\times \left(4i\right)-4i\times 2-4\times 4\left(-1\right).

$$Re(-6-12i-8i+16)$$

Combine the real and imaginary parts in -6-12i-8i+16.

$$Re(-6+16+\left(-12-8\right)i)$$

Do the additions in -6+16+\left(-12-8\right)i.

$$Re(10-20i)$$

The real part of 10-20i is 10.

$$10$$